Say we have an AF C* algebra $A$ described by some Bratteli diagram $E$. If $M_\infty (A)=\displaystyle{\lim_\rightarrow M_n(A)}$ and $P(A)$ are the projections in this algebra, we know that $K_0(A)^+=P(A)/\sim$ where $\sim$ is the von Neumann equivalence relation, and the Grothendieck group construction gives us $K_0(A)$.

My question is: is there an analogous description of projectors and equivalence at the level of the graph $E$ itself which yields the same $K_0$ group as the AF algebra it describes? I have thought about this question in terms of the path algebra of $E$, but really don't know much about this area.

1 Answer
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Kerov's thesis "Asymptotic representation theory of the symmetric group" has a nice discussion of these ideas. I think I can remember an answer to your question in that book, but I'll have to check later.

As you probably know, the ordered abelian group $K_0(A)$ is a complete invariant of the algebra $A$. Here is a construction that recovers not only $K_0(A)$, but the positive cone as well.

Consider a certain class of integer-valued functions on $E$ which obey the same branching rule as the dimension function:
$$ f : E \longrightarrow \mathbb{Z}$$
$$ \sum_{\lambda \nearrow \Lambda} f(\lambda) = f(\Lambda),$$
Where the notation $\lambda \nearrow \Lambda$ indicates that $\Lambda$ covers $\lambda$ in the Bratteli diagram.
We don't actually require $f$ to be defined on all of $E$, just almost everywhere (past some finite stage). The condition is only required to hold in cases where $f$ is defined.

Such functions form a group under addition. The nonnegative cone is functions which are nonnegative in their domain. I think it is straightforward to check that this construction produces $K_0(A)$ as an ordered abelian group.